Problem: Let $g(x)=\sqrt{5x-1}$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $[1,10]$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2.25$ (Choice B) B $4.25$ (Choice C) C $6.5$ (Choice D) D $8$
According to the Mean Value Theorem, there exists a number $c$ in the open interval $(1,10)$ such that $g'(c)$ is equal to the average rate of change of $g$ over the interval: $g'(c)=\dfrac{g(10)-g(1)}{(10)-(1)}$ First, let's find that average rate of change: $\dfrac{g(10)-g(1)}{(10)-(1)}=\dfrac{7-2}{9}={\dfrac{5}{9}}$ Now, let's differentiate $g$ and find the $x$ -value for which $g'(x)={\dfrac{5}{9}}$. $g'(x)=\dfrac{5}{2\sqrt{5x-1}}$ The solution of $g'(x)=\dfrac{5}{9}$ is $x=4.25$. $x=4.25$ is indeed within the interval $(1,10)$. In conclusion, $c=4.25$.